should the `black_box` asm block be marked volatile? about subtle HOT 5 CLOSED

I guess this part, 4 min later, is maybe more relevant: https://www.youtube.com/watch?v=nXaxk27zwlk&amp=&t=44m2s

"the compiler needs to know that because the compiler, my compiler, has an assembler built into it. it's going to take a look at that string of inline assembly, and say, i know how assembly works. i can do that. it's going to assemble it, and build a representation for it, and then it's going to notice that it doesn't do anything. or in my case, that it's empty. that doesn't do what I need... so in my opinion, this is a correct usage of volatile, and a correct usage of inline assembly."

from subtle.

I'm pretty sure the black_box here is based on the implementation from libcore:
https://github.com/rust-lang/rust/blob/342c5f33d097b2dc07a2dbc0ca45a37379d2ff60/src/libcore/hint.rs#L109

So feel free to also comment on the related tracking issue: rust-lang/rust#50297

from subtle.

It looks to me that right now it doesn't really matter. but it might matter more in the future? I don't know enough about llvm to answer that.
https://godbolt.org/z/D3j7rf

from subtle.

I was reading the gcc asm docs, it says this: (https://gcc.gnu.org/onlinedocs/gcc/Extended-Asm.html#Volatile) on the first line

GCC’s optimizers sometimes discard asm statements if they determine there is no need for the output variables. Also, the optimizers may move code out of loops if they believe that the code will always return the same result (i.e. none of its input values change between calls). Using the volatile qualifier disables these optimizations. asm statements that have no output operands, including asm goto statements, are implicitly volatile.

So maybe the statement is implicitly volatile because it has no output operands?

from subtle.

closing, thank you!

from subtle.